Voltage drop from R/R output to battery terminals

[Squishy]

Hello all,

I've been having this 'problem' since I have my bandit and decided to try and find the source.
I've become quite familiar with my 400 but don't have that much experience with electronics.

So, with engine running, I'm only measuring about 13.2V charging volt on the battery. At 5000rpm, it goes down to 12.5V.
My AC generator is producing a healthy 75V+ AC at 5000rpm. (as per manual)
When I measure right at the output of the R/R (everything is still connected and engine running) I can see a much better 14V+!

My conclusion is that I'm losing voltage across a wire or connection from the R/R to the battery.
If I measure the voltage drop across the + output from the R/R and the + on the terminal of the battery, I can see 0.5V drop.
The same goes for - output R/R to - terminal battery, for a total of 1V.

Now my question is, is my conclusion correct?
On the wiring diagram I can see that the ground wire has a very short path and directly connects to the main ground wire that connects to the battery ground terminal.
The + wire path is also pretty short and can go directly via the fuse box to the + terminal on the battery.

Isn't this the path it will take (least resistance?)? Shouldn't the voltages be roughly the on the output R/R and the battery?
Or is the voltage drop logical because of all the loads on the circuit 

Thanks

[ventYl]

Electrical circuit schematics do not reflect real wire / path topology. The fact that some path is short in schematic does not automatically mean that same path will be physically short.

If you are able to measure some voltage between output of R/R and + of Battery then definitely this path has some resistance causing voltage drop of 0,5 / 1V. Usually if insulation on your wires does not have signs of rupture or is not falling off the resistance (and in turn voltage drop) is caused by contacts which are corroding over time and developing "transfer" resistance. This also makes sense with larger voltage drop reading on ground path while both battery - and wiring grounds are connected to chassis. These connections tends to corrode by electro-chemical corrosion as the place where two different metals (here very roughly copper and steel) have contact and electrical current is passing this contact. (btw: The direction in which current is passing is significant, which is reason why some vehicles back in mid 20th century tend to have grounded +... corrosion is weaker when current is passing opposite direction)

While contacts on positive path are mostly copper - copper these only have signs of normal copper corrosion and most probably developing smaller resistances (or these contacts are opened more often thus corrosion is scratched). Under some circumstances you should be able to measure resistance of the path itself (if there are no loops in schematics and your meter is able to measure low resistance with low error).

[Squishy]

Yes I know the lengths used in the drawing aren't to scale. But it does show the ground wire of the R/R is connected directly to the ground wire black/white ground wire of the battery, which I assume can't be long because it connects to the main ground wire in the wiring harness (Wouldn't make sense to take it all the way to the front?).

Anyway I checked all cleaned all battery terminals and connections. I also checked the connector of the R/R but I'm losing no voltage over the connecter at all.
The voltagedrop is somewhere in the wiring harness.

The resistance on the wire that has the voltage drop is practically the same as direct connection of the probes (0.07 ohm). But of course this is because there's no load. Under no load there's no voltage drop either.

The only thing I can't wrap my head around is how the current will flow and whether all the branches of the wiring harness have any affect on the voltage.
The wiring harness looks to be in good condition from the outside and I'd rather not open it up if I don't have to.

Perhaps somebody has the opportunity to compare their voltage when measured with the probes right into the output connector of the R/R and vs. at the terminals (running engine).

[ventYl]

The main problem with such a measurement is that basic cheap chinese electronic multimeters have high error rate when measuring small resistances when switched to ohmic reading. Indirect reading (like using Ohm's law, measure current through wire using known(!) source) will be more accurate.

You can try to measure voltage drop across connector if both ends of connector have unisolated wires visible (maybe will be visible if you pry the outer insulation a little bit). You can then measure voltage drop across sections (like across R-R connector ends, across battery connector ends, wire section between connectors) and then tell which section does have far highest part on voltage drop. Absolute resistance values do not have any meaning here while every machine has 20 years of completely different history.

[Squishy]

Yeap as I said there's no voltage drop across the connector.
The whole 0.5V drop is across the wire that comes out of the wiring harness to the battery ground terminal, and the output of the the R/R.

The problem is I can't really put a known source on the wire because it needs to have a load for any measurement to make any sense. Worn wires will show their bottleneck only when there's a load on them.

I just think there is something else causing the voltage drop...
I am not sure whether the voltage should be the same across the battery and across the R/R even if the wires were perfect... That's what I'm trying to figure out.

[ventYl]

Well yes the voltage should be the same at both ends of wiring if wiring was perfect without any resistance. You can imagine as if the wiring was replaced by resistor of resistance you've measured connected to battery and then another on ground connection. Battery itself is resistance for generator too. The more resistance the more charge it holds. Then you have series connection of three resistors. Voltage loss at resistor is proportional to it's resistance (U=R.I - Ohm's law).

If you don't understand my hard-to-follow basics of electronics then the only thing you need to carry from above is that if there's voltage drop, then there's resistance and if the resistance won't be there no voltage drop ocurred.

[Chris H]

After going through 3 regs in a year I looked into this same problem and after reading the mods done by the SV guys treated the bike to a cbr6 reg from the 2008 model (F008).
If you google sv reg mod you will find loads of info on the sv forum regarding alternative regs and wiring mods, they list better connections and direct wiring, everything you need is on there.

[Squishy]

After going through 3 regs in a year I looked into this same problem and after reading the mods done by the SV guys treated the bike to a cbr6 reg from the 2008 model (F008).
If you google sv reg mod you will find loads of info on the sv forum regarding alternative regs and wiring mods, they list better connections and direct wiring, everything you need is on there.

Thanks,
I've thought about direct feed to battery but that would skip the fuse and key switch... so figured it wasn't a good idea.
edit: I've read a bit and the guy introduces a new 30A fuse for it.
It sounds like a good idea but I wonder what path the original wiring takes? If you look at the wiring diagram surely the stock setup can't be that bad unless the wires are all bad?

[ventYl]

in cars; the path from generator (regulator) to battery is usually unfused and without any switches. while rectifier is able to hold charge in battery ensuring it won't flow into alternator wiring this circuit does not have to be switched.

it is nothing else that generally a bad idea to put fuses and/or switches into line where potentially high currents can flow (400W @ 12V => ~ 30A).

[Squishy]

in cars; the path from generator (regulator) to battery is usually unfused and without any switches. while rectifier is able to hold charge in battery ensuring it won't flow into alternator wiring this circuit does not have to be switched.

it is nothing else that generally a bad idea to put fuses and/or switches into line where potentially high currents can flow (400W @ 12V => ~ 30A).

Sorry can you rephrase that last sentence? Are you saying it's a bad idea to put a fuse between the R/R and the battery on a direct connection?
Where did you get the 400W from?

[Chris H]

The original path is not too bad, problems mostly come from slightly light weight original wires connected using the cheap and nasty spade connectors, these fill up with water and then load the reg,rec.
You either get a blown reg or melted connector.

The SV uses the same model as the B4 so they are the exact problems we have.
Since doing the F008 conversion I'm getting 14.3 volts no matter what.

[ventYl]

in cars; the path from generator (regulator) to battery is usually unfused and without any switches. while rectifier is able to hold charge in battery ensuring it won't flow into alternator wiring this circuit does not have to be switched.

it is nothing else that generally a bad idea to put fuses and/or switches into line where potentially high currents can flow (400W @ 12V => ~ 30A).

Sorry can you rephrase that last sentence? Are you saying it's a bad idea to put a fuse between the R/R and the battery on a direct connection?
Where did you get the 400W from?

Yep I am saying that it is generally bad idea to put anything else than huge direct wire (fuse, switch, portal, etc.) between R/R output and battery. 400W was just a guess (but GSF1200 '99 has 405W @ 5000 RPM so GSF 400 might be similar?). As least pessimistic boundary you can get a sum of all loads which together give you 60+5+5+21+50+50+60=251W (high beam + tail light + park light + brake light + turn light * 2 + indicator + ignition + fan - let's say normal draw during city ride with DRL on warm day...). You definitely don't want your battery get drained under these conditions so power will be greater than this. If only by 50 watts then the resulting power is 300W. Whole this power has to be transferred from R/R into battery by single wire and only then to be distributed to various draws.

If you use formula for computing power (P=U.I -> I = P/U) then 300/14.4 = 20.8 A of total flow (14.4 is your target voltage). Ofc actual numbers will be lower because you are not using all available output at all times but you have to set the wiring to be able to sustain these conditions. With nearly 20A passing through the wire if you put there connector with resistance of 0.1 Ohm, your voltage drop will be (U=R.I) 20*0.1 = 2V. Ofc contacts do not have such a high resistance but you don't have to have drop of more than let's say 0.2V on each wire (0.5V combined) otherwise your regulated 14.2V will make some 13.7V at battery terminals which is lower bound for healthy charging of car batteries (which are similar to motorcycle batteries thus I assume that same conditions hold). This limits you to resistance of not more than 0.01 ohm per whole line or roughly 0.005 ohm per contact (in case of zero resistance of wire itself which is purely ideal; in real cases wires do have some resistance). You stated that your voltage drop is 0.5V at positive side. This means that cumulative resistance of connection between R/R and battery under above conditions couldn't be larger than 0.025 ohm (R=U/I; 0.5V/20A). In real case the boundary is little bit larger because of the current passing through wire would be lower.

You stated that your cheapo multimeter shows base resistance of sole probes as 0.7 ohm. That's one order of magnitude larger than resistance you are searching for (even if your current was half of my sample computations the resistance is still only 0.05 ohm). Either you have some poor connection in probe wiring or cheapo multimeter is not able to get correct 0 reference. In latter case we cannot expect it to be able to measure 10^-2 resistances accurately. In former case the resistance we want to measure is 1/10th of parasite resistances we are dealing with. Not a good situation to perform accurate measurement (but I bet that the latter case holds for you now).

With fuse situation is similar (each fuse ~ 2 contacts) with additional problem of how "large" fuse to use. With fuses the problem is that they don't vapour magically at the time the current passing through them reaches the number written at their package. Heat has to melt the connection in fuse. Generally the numbers are: fuse nominal current should melt it in under 60 seconds, 10x the nominal current should melt it under 1 second.

Usually if something breaks in regulator then regulation (shunt) path in R is opened (not shortened) and your installation gets full voltage of regulator only lowered by "softeness" of alternator (basically internal resistance of alternator wirings which says how high the voltage output of sole alternator will be when some defined load is attached). While overvoltage resistances are pretty much the same (or lower if some bulb breaks) thus fuse would not get any use here (because fuse is melted by heat; while amount of heat dissipated by flowing current is defined by P=R.I²; You can see that voltage is not present in the formula thus is completely irrelevant) so fuse won't react.

In case of short circuit between R/R and battery the current flowing through short circuit would be larger than nominal 30A and momentarily maybe even larger than 150A (defined by short-circuit current of your battery, usually more than 10x of nominal capacity in Ah with healthy battery) but it won't be generated by alternator. It would be sourced from battery because with such a low resistance voltage output of regulator would drop to some 6-9 volts and current would be limited to somewhere around 60A. The time in which fuse would react for such overcurrent is somewhere around 15 seconds. The fuse would react for such a case if and only if the connection was between R/R and fuse allowing to pass current from battery through fuse. Even in such case current would be supplied to short-circuited connection even after fuse had blown. Well theoretically. Practically the voltage drop on battery due to short circuit would drop system voltage to so low values that ignition stopped to work (usually the lower boundary for all kinds of ECU and CDIs is 9V) and engine halted immediately eventually damaging CDI.

Above shows that any kind of fuse is purely useless in this line and that's the reason why no fuse is there. All you want is the best, bulkiest wire you can fit in with best connections you can manage. Nothing more, nothing less. Additionally change of actual numbers compared to theoretical ones used in this post wouldn't make a huge difference I think. Change of generator power by 100W is change of about 30%, current by 10A makes 50%. Still not enough to make change one order of magnitude large, something which would make noticeable difference in calculations above.

[Squishy]

Thanks for that.
I guess when overvolt is the problem then yeah a fuse is not gonna help much.
If 20A is passing through the b/w and red wires from the R/R then I find the stock ones pretty small for their task...

Btw I said 0.07 ohm not 0.7 ohm on the wire.

[ventYl]

there are formulas to compute required wire section surface for good (low) resistance and good current pass (the driving factor here is current density). while Suzuki most probably used them when designing bandit's wiring the coefficients in those formulas hold for brand new wires. Even if insulation on wires looks good it is possible that humidity was dragged deep into wiring due to capillary forces.

In cars the wire from generator to battery is usually the same/similar thickness as the wire from battery to starter. With bikes there's no need/place for such a thick wire (generator is much less powerful and weight is a concern) but still has to fulfill some specs.

[Squishy]

OK I'm going to clean all connectors and check the wires and if it doesn't improve the charging volts I will consider doing a direct wire mod.
Will report back later.